I am trying to design a WWI propeller so I started calculating. The propeller comes with a specified pitch of 1.75m. The engine is supposed to run at around 1900 rpm and 147kW. These are most likely not the maximums, but I use these values as a starting point.

So I calculated the propeller speed: 1900/60 * 1.75 * 3.6 = 199.5 km/h

I also found on the internet an angle of 16.88 degrees but the site mentions "angle of attack" which I deem incorrect. However if I treat the value as "effective pitch angle" I get about the same speed of ~195.68 km/h (using the formula from Koyovis for Φ from How to calculate variable pitch propeller parameters?).

Everything looks good, my reasoning at this point is that this prop is suitable to be used for the particular aircraft. The speed that I found seems to be reasonable for this aircraft.

But here comes the problem, this is the part that I don't understand and that is 'thrust': if the propeller reaches 1.75 pitch this means that the exit velocity - the input velocity (Ve - V0) is 0!

If I understand the formulas (such as here) correctly this means there can no longer be thrust. How would it possible then that the aircraft is still being pulled forward? This automatically means the aircraft has to slow down. But my assumption is that the propeller is 'designed' to go as fast as the specified pitch.

Can someone enlighten me please. Why does my propeller have to be slower than designed?

edit (I will leave my question above unaltered). Clarification as per request of JZYL:

The propeller is: AB622 Some of these propellers can still be found on auction sites and such. They have markings like "G 1669 NO 90 DRG AB 662 CRH HISPANO SUIZA, WOLDSLEY VIPER D2400 M/M p 1750 2400mm"

The pitch is 1750mm The diameter is 2400mm

The aircraft I am looking at is a WWI SE5a with a Wolseley W.4A Viper engine. It is difficult to state exact parameters. There is a lot of 'belief' of what the aircraft at that time could do. Some people say the aircraft top speed would be 138 mph (220 km/h). However might as well be one top speed ever recorderd for one plane at excellent flying contditions. Also some people claim the engine could run at 2200 RPM.

So instead I go for some 'safe' or nominal values: Speed = 125 mph RPM = 1900

The calculations that I make do not add up. The "pitch speed" (I don't know what the official term is) that I have calculated for the prop is 199.5 km/h (124.6 mph) with these parameters.

But if I have to take slip of say 10% into account (how much is normal, 5% 10%?). In that case I go down to: 112 mph.

112 mph is way below what people see as the maximum speed of the craft.

So I am basically trying to find a clarification for this.


1 Answer 1


The geometric pitch ($p$) is simply the distance that a point on the prop would move forward in one rotation, if it's gripping in a solid medium. You can derive the pitch speed via this geometric relation if you wish, but I would caution against inferring performance from this number. Instead, you should consult propeller, engine and aircraft performance in their entirety.

Using your provided data, let's see if everything fits. We assume:

  • Prop RPM of 2200
  • Top cruise speed ($V$) of 125mph
  • Prop diameter ($D$) of 2.36m (from Wikipedia on SE5a)
  • Prop pitch ($p$) of 1.75m
  • Wing area ($S$) of 22.7 sq.m (from Wikipedia on SE5a)

From this reference, the reference pitch angle/blade angle ($\theta$) by convention is measured at 3/4 of the blade radius, and can be related to pitch via:

$$\theta=\tan^{-1}{\frac{4}{3} \frac{p}{\pi D}} \approx 17.5deg$$

The advance ratio is calculated, with $n$ being the prop rate of rotation in round per seconds:

$$J=\frac{V}{nD} \approx 0.65$$

Using the plot below as a rough reference, which is predicted for the McCauley 7557 propeller, gives a thrust coefficient ($C_T$) of 0.06.

Prop data

Ref: http://webserver.dmt.upm.es/~isidoro/bk3/c17/Propellers.pdf

Now the thrust coefficient is defined as:

$$C_T=\frac{T}{\rho n^2 D^4}$$

where $rho$ is the air density and $T$ is the propeller thrust.

For level flight, thrust equals drag, and we can back calculate the drag coefficient:

$$C_D=\frac{2C_T n^2 D^4}{V^2S} \approx 0.07$$

This is high enough to be well within the realm of possibility for a biplane (if it's too low, then it's an indication that the performance numbers do not jive). For reference, Cessna 172 has a total drag coefficient of 0.032 at 122kt at SL. In conclusion, I would say that your numbers check out in the back-of-envelope sense.

  • $\begingroup$ I assumed no such thing. I clearly said that: "This automatically means the aircraft has to slow down". In other words: "The aircraft is slower". And this is what I am trying to find a clarification for. The aircraft speed then becomes much slower than what the books say. $\endgroup$
    – Veltro
    Apr 7, 2020 at 18:51
  • $\begingroup$ @Veltro Then I have misunderstood your question. I suggest you modify your OP, cite what you believe the aircraft speed is, all the factual info, etc. $\endgroup$
    – JZYL
    Apr 7, 2020 at 18:54
  • $\begingroup$ @Veltro Please see my modified answer. $\endgroup$
    – JZYL
    Apr 7, 2020 at 23:32
  • 1
    $\begingroup$ Your CT looks a bit high (0.06 instead of 0.05). RPM of 2200 is too high compared to other sources I found (RPM of 2140 absolute max). Anyways I ran the formula's in my excel sheet. Eventually I used RPM 1945 D 2400 V 125 mph (Choosing the values carefully to get a Φ of ~16.88). J becomes 0.71. Lookup of CP 0.039 CT 0.046. Then calculated CD 0.04578. Compared to the CD found for the aircraft 0.0402 (zero lift) to 0.0440 (max speed) this value comes really close but still lower. So I guess the numbers 'jive'. Thanks! Now going to figure out how to calc the thrust.and if the engine matches. $\endgroup$
    – Veltro
    Apr 8, 2020 at 12:33
  • $\begingroup$ @JZYL V/n in advance ratio formula is geometric pitch or aerodynamic pitch/"effective pitch"? $\endgroup$
    – user52248
    Sep 16, 2020 at 18:44

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